Do electrons coming out of a lightbulb (and going back into the circuit) slow down?
The electrons enter the light bulb filament with relatively high kinetic energies. As they travel through the filament they collide with metal atoms transferring much of their kinetic energy to the metal. This energy raises the temperature of the metal. The metal in turn radiates this energy as electromagnetic waves, many in the visible spectrum.(Source 1)
Each light bulb results in a loss of electric potential for the charge. This loss in electric potential corresponds to a loss of energy as the electrical energy is transformed by the light bulb into light energy and thermal energy. (Source 2)
My understanding of the above sources is that after the electrons come out of the lightbulb, they have less electrical potential (Voltage) than when they entered the lightbulb (as per Source 2).
Does this mean the electrons travel slower (as they have lost kinetic energy) between the bulb and the positive terminal of the battery (compared the the negative terminal of the battery to the bulb)?
Does this in term mean the current (rate of flow of charge) is less in the 2nd half of the circuit (i.e. between the bulb going towards the positive terminal) compared to the first half (between the negative terminal and the bulb)?
If not, what is the actual difference between the electrons coming into the bulb and going out of the bulb back into the circuit? If you could explain the difference in terms of terms voltage/current/charge but also what physically happens to the electrons e.g. do they travel faster, slower, that would be useful.